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PhoenixFire

  • one year ago

\[F(x)=\int _{ 1 }^{ ln(x^{ 2 }+e) }{ sin(e^{ t }) } dt\] Find F'(x). Not sure how to do this, I understand from the Fundamental Theorem of Calculus that \[F'(x)=\frac { d }{ dx } \int _{ a }^{ x }{ f(t) } dt=f(x)\] But what happens to the limit of \(ln(x^2+e)\)?

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  1. bahrom7893
    • one year ago
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    Don't you just do this: Sin(e^b)*(Sin(e^b)) ' where b is the upper limit.

  2. bahrom7893
    • one year ago
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    Well, you'd then subtract Sin(e^1) * (Sin(e^1))', but since the derivative of a constant is 0, this term becomes 0, and you're left with just the first one.

  3. bahrom7893
    • one year ago
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    Oh and by the way, e^(ln(x)) = x

  4. PhoenixFire
    • one year ago
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    I don't understand why you're multiplying f(x) by f'(x) Or do you do the integration, and then derive it? \[\frac{d}{dx}(sin(e^{ln(x^2+e)})-sin(e^1))=\frac{d}{dx}sin(x^2+e)=2xcos(x^2+e)=F'(x)\]

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